In the rectangular coordinate system above, if OP < PQ, is the area of

In the rectangular coordinate system above, if OP < PQ, is the area of region OPQ greater than 48 ?

(1) The coordinates of point P are (6,8). Now, if OP were equal to PQ then the triangle OPQ would be isosceles and OS would be equal to SQ and the area would be: 1/2*base*height=1/2(OS+SQ)*PS=1/2*(6+6)*8=48. Since OP<PQ then OS<SQ and the base OQ is more than 12, which makes the area more than 48. Sufficient.

(2) The coordinates of point Q are (13,0) --> we know the length of the base (OQ=13) but know nothing about the height (PS), which may be 1 or 100, so the area may or may not be more than 48. Not sufficient.

Answer: A.

Hope it's clear.


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thanks Bunuel.
stmt 1 shows us that the smaller triangle is 1/2*6*8 and the other triangle must be larger than the smaller one. adding the two triangles together you get a number larger than 48.

Hi Bunuel,

Can thehre be a case when OS= 5.5, SQ=6.5? Then OQ= 12 ....

No, that's not possible. We know that the coordinates of point P are (6,8). PS is altitude, thus the coordinates of point S are (6,0), so OS=6.

Hope it' clear.
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Hi,

Could you explain the following :

Since OP<PQ then OS<SQ and the base OQ is more than 12


Thanks.

Since OS=6 and OS<SQ, then SQ>6, thus OQ=OS+SQ=6+(more than 6)=(more than 12).

Hope it's clear.
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Hi Bunuel,
Is there any other way(as fast as the given solution) to solve this without drawing the perpendicular from P to X-axis?
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I didn't understand how are we assuming that OS is equal 6? and how are we concluding t either that OS=QS?Can you pls explain?

Please read the thread: in-the-rectangular-coordinate-system-above-if-op-pq-is-129092.html#p1244697

The coordinates of point P, which is just above S, are (6,8), thus the coordinates of point S are (6, 0).
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Hi Bunuel,

Option (2)
Isn't it possible to get the answer, if OPQ>48, from...

1) OP<PQ
2) a^2+b^2=13^2 .... (a=OP, b=PQ)

I think there is no option of (a,b) to make ab>96 when OP<PQ because a^2+b^2=13^2 and ab=96 can not have any common answers.

a^2 +(96/a)^2 = 169 ...........ab=96
a^4-169*a^2+8716=0
A^2-169A+9216=0................A=a^2
169^2 -4*9216 = -6303............b^2-4ac <0 (ax^2+bx+c=0)

Thanks.

This is not correct because we are not given that OPQ is a right triangle, thus you cannot write OP^2 + PQ^2 = OQ^2.

Hope it's clear.
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From St 1, we have the co-ordinates of P (6,8) ie. Pt S (6,0) and Pt O (0,0)

Now let's take SQ=x so basically the questions becomes is 1/2 (6+x)*8>48

or (6+x)>12 or x>6

We know OP =10 so PQ>10.....If now we take PQ= 10.5 and apply Pythagorus

(10.5)^2= 8^2+x^2 -------> x^2= 110.25-64 = 46.25 ie x =6.8 which is more than 6 and therefore sufficient

Option B,C AND E ruled out.

From St 2, we know only OQ =13 but don't know the height so not sufficient.

D ruled out so Ans is A
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Although I understand the discussed solution I come to to a different solution due to a different interpretation of the question. Maybe some of you have a similar opinion...

In my opinion, the question at no point requires point Q to lie on the x-axis. The question only requires OP<PQ. Although the graph and question may be interpreted in a way that leads you to think that Q lies on the x-axis they certainly don't state this requirement explicitly.

If OP has a length of 10 and PQ has a length of 10+ but Q is located not on the x-axis but rather close to the origin the area of OPQ may be less than 48 despite having P located at (6,8)

I attached a graph of my example.

I know that the solutions manual offers the same solution as the forum agrees upon here. Is the solution manual wrong? Please point out any mistakes of mine.

Hi Bunuel,

This is my first post.

Can you please explain how we got OS<SQ?

When I solved, I took the smallest value of Q possible i.e. Q (7,0) and got Area= 1/2 *7*6 = 21 so concluded that both (i) and (ii) are required.

Hi robinpallickal

You are probably overlooking the condition given in the question that OP < PQ

If \(OP < PQ\)
-> \(\sqrt{OS^2 + PS^2} < \sqrt{SQ^2 + PS^2}\)
-> \(OS < SQ\) (as OS and SQ are both positive)

Therefore, you can't take the x-coordinate of Q as anything less than 12.
Hope that clears your doubt.

So, we have OP<PQ, Area of OPQ = (PS*OQ) / 2 > 48 --> PS*OQ > 96 ?

1) P (6,8) means --> Heght PS=8, OS=6 then we can find OP^2=8^2+6^2=100, OP=10
Since PQ>OP --> SQ>6 (SQ^2=PQ^2 (let's say 10,1) - 64 means SQ>6) --> OQ>12 -> 8*OQ(>12) > 96 SUFFICIENT
2) We need the height to be able to calculate the area - INSUFFICIENT

Hi Bunuel,

Could you explain why OS < SQ? is there a theory/concept behind it?

PS^2 + OS^2 = OP^2
PS^2 + SQ^2 = PQ^2

Since we are given that OP < PQ, then OS < SQ.

Hope it's clear.
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Let us say OP=PQ
1) P= 6,8
OP =10
PQ=10 we have assumed
=> Q = 12,0

A= 0.5 x 8 x 12 =48
But it is given OP < PQ => A<48 DEFINITE NO
Sufficient
2) Q= 13,0
Insufficient as it will depend upon height ie y coordinate of P
if y coordinate of p < 96/13 then only above false => if y coord of P > 96/13 above is true
we dont know y coord of P => INSUFFICIENT


ANSWER IS A